JEE Physics Motion In A Plane Notes Set 05

Projectile Motion Basics

1. A projectile moves under gravity alone. The horizontal and vertical motions are independent: horizontal velocity \(u\cos\theta\) stays constant, while the vertical motion has acceleration \(-g\) starting from \(u\sin\theta\).
2. At time t: \(x=u\cos\theta\,t\), \(y=u\sin\theta\,t-\tfrac12gt^2\), \(v_x=u\cos\theta\), \(v_y=u\sin\theta-gt\). The path is a parabola, \(y=x\tan\theta-\dfrac{gx^2}{2u^2\cos^2\theta}\), which can also be written \(y=x\tan\theta\left(1-\dfrac xR\right)\).
3. The velocity is perpendicular to the acceleration only at the highest point. The magnitude of the rate of change of velocity is always g.
4. At equal heights the speed is the same, and the vertical velocity just changes sign.

Ground to Ground Formulas

For the same speed, angles \(\theta\) and \(90^\circ-\theta\) give equal ranges, while their heights are in the ratio \(\tan^2\theta:1\) and \(h_1h_2=R^2/16\). Doubling the speed multiplies the range by four. The kinetic energy at the top is \(K\cos^2\theta\), and the change in momentum between launch and landing is \(2mu\sin\theta\) (vertical). The highest point of the path as seen from the launch point has elevation \(\tan\alpha=\tan\theta/2\).

Horizontal Projection and Height Problems

1. A body thrown horizontally with speed u from height h: time of fall \(t=\sqrt{2h/g}\), horizontal range \(u\sqrt{2h/g}\), and displacement \(\sqrt{u^2t^2+\tfrac14g^2t^4}\).
2. At time t the velocity makes an angle \(\phi\) with the horizontal where \(\tan\phi=gt/u\).
3. For a throw from a height (or onto a lower roof), write \(y=u_yt-\tfrac12gt^2\) with the final y negative and solve the quadratic for t; reject the negative root.
4. For a jump across a gap: minimum speed equals gap divided by the time to fall the height difference.

Relative Velocity in Two Dimensions

1. \(\vec v_{AB}=\vec v_A-\vec v_B\). Perpendicular velocities of equal magnitude v give a relative speed \(v\sqrt2\).
2. Two bodies move towards a collision only if their relative velocity points along the line joining them.
3. For the closest approach, work in the frame of one body and find the point where the relative position is perpendicular to the relative velocity.
4. Pursuit problem: n persons at the corners of a regular polygon of side d, each heading at the next, meet after \(t=\dfrac{d}{u(1-\cos(2\pi/n))}\).

Rain and Man

1. Rain falling vertically with speed v and a man walking at u: the rain appears to come at angle \(\tan^{-1}(u/v)\) from the vertical, with relative speed \(\sqrt{u^2+v^2}\).
2. The umbrella or tube must be tilted from the vertical by this same angle, towards the direction the man is moving.
3. To make rain appear vertical, the man must move with the horizontal component of the rain velocity.

River and Boat

Here w is the river width, \(v_b\) the boat speed in still water and \(v_r\) the river speed. The resultant speed when heading straight across is \(\sqrt{v_b^2+v_r^2}\). The angle made with the flow when going straight across is \(90^\circ+\alpha\).